A) n
B) 2n
C) pn
D) \[{{p}^{n}}\]
Correct Answer: B
Solution :
[b] If \[N={{p}_{1}}^{{{\alpha }_{1}}}{{P}_{2}}^{{{\alpha }_{2}}}\] than the sum of the divisors of N is \[\left( \frac{{{p}_{1}}^{{{\alpha }_{1}}+1}-1}{{{p}_{1}}-1} \right)\left( \frac{{{p}_{2}}^{{{\alpha }_{2}}+1}-1}{{{p}_{2}}-1} \right)\]You need to login to perform this action.
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